Lie Groups and Their Representations

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These notes are being transcribed, condensed, and illustrated. They may lag or omit details, please refer to the handwritten notes for the full lecture.

This note introduces Lie groups — continuous groups parameterized smoothly by a finite set of parameters — and shows how the tools we developed for finite groups (similarity transforms, direct sums, tensor products, decomposition, QS = RQ) carry over to Lie groups by working with generators instead of group elements. We then apply these ideas to derive the irreducible representations of \mathrm{SO}(2) and \mathrm{SO}(3).

1) From Finite to Infinite Groups

So far we have worked with finite groups: the group has a finite number of elements, we can list them all, and we can build the multiplication table explicitly. But many symmetry groups in physics are infinite — the rotation group \mathrm{SO}(3) has continuously many elements parameterized by rotation angles.

Generally, infinite groups are hard to work with. But a special class called Lie groups are particularly nice: their elements are parameterized smoothly by a finite set of continuous parameters.

g(\vec{\alpha}) \in G, \qquad \vec{\alpha} = (\alpha_1, \alpha_2, \ldots, \alpha_N)

Similar parameters \Rightarrow similar group elements. Examples include \mathrm{SO}(3) (3D rotations, 3 parameters), \mathrm{SU}(2) (complex 2D rotations, 3 parameters), and \mathrm{SL}(2, \mathbb{R}) (real 2 \times 2 matrices with determinant 1, 3 parameters).

1.1) The Exponential Parameterization

Like all groups, Lie groups have an identity element g(\vec{0}) = e with representation D(\vec{0}) = \mathbb{1}.

Infinitesimally away from the identity, we can Taylor expand:

D(d\vec{\alpha}) = \mathbb{1} + (d\alpha_a) X_a + \ldots \qquad \text{(Einstein summation)}

where X_a are the matrix generators of the group:

X_a = \frac{\partial D(\vec{\alpha})}{\partial \alpha_a} \bigg|_{\vec{\alpha} = 0} \qquad \text{for } a = 1, \ldots, N

For finite (not small) \vec{\alpha}, we can build up a group element from many infinitesimal steps:

D(\vec{\alpha}) = \lim_{k \to \infty} \left( \mathbb{1} + \frac{\alpha_a X_a}{k} \right)^k = e^{\alpha_a X_a}

This is the exponential parameterization. The matrix exponential e^X = \sum_{k=0}^{\infty} \frac{1}{k!} X^k is always well-defined, and e^X is always invertible (its inverse is e^{-X}).

Key idea: While a Lie group has infinitely many elements, it has only finitely many generators. All the information about the group is encoded in the generators and how they combine.

1.2) The Baker-Campbell-Hausdorff Formula

When combining two exponentials of matrices, the result is not simply the exponential of their sum (unless the matrices commute). The Baker-Campbell-Hausdorff (BCH) formula gives:

e^X e^Y = e^Z, \qquad Z = X + Y + \frac{1}{2}[X, Y] + \frac{1}{12}[X, [X, Y]] - \frac{1}{12}[Y, [X, Y]] + \ldots

where [X, Y] = XY - YX is the commutator of X and Y. The commutator measures how different XY is from YX.

The Lie algebra condition: Since e^Z must also be a group element, Z must be a linear combination of generators. This means the BCH series — which involves only commutators of X and Y — must close on the generators. In other words, the commutator of any two generators must itself be a linear combination of generators:

[X_a, X_b] = f_{abc} \, X_c

These coefficients f_{abc} are called the structure constants of the Lie algebra. They encode all the information about the group's structure.

Properties of the commutator:

Bilinearity: [aX + bY, cZ] = ac[X, Z] + bc[Y, Z]
Alternativity: [X, X] = 0
Anticommutativity: [X, Y] = -[Y, X]
Jacobi identity: [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0

2) Deriving Generators for \mathrm{SO}(n)

Elements of \mathrm{SO}(n) are n \times n orthogonal matrices with determinant 1:

R^T R = \mathbb{1}, \qquad \det(R) = 1

If R = e^A, what constraints does the orthogonality condition impose on the generator A?

Step 1: The transpose of the matrix exponential is the exponential of the transpose:

(e^A)^T = e^{A^T}

This follows from the fact that transposing a product reverses the order: (ABC)^T = C^T B^T A^T. When all factors are the same matrix, the reversal doesn't matter: (A^n)^T = (A \cdot A \cdots A)^T = A^T \cdot A^T \cdots A^T = (A^T)^n. Applying this term by term in the power series e^A = \sum \frac{1}{n!} A^n gives the result.

Step 2: The orthogonality condition R^T R = \mathbb{1} becomes:

e^{A^T} e^A = \mathbb{1}

Step 3: The condition e^{\alpha A^T} e^{\alpha A} = \mathbb{1} must hold for all values of \alpha (since \alpha A is also a valid generator). Since it holds for all \alpha, its derivative must also vanish. We evaluate at \alpha = 0 (the identity element) because the exponentials collapse to \mathbb{1}, leaving a clean linear constraint on the generator:

\frac{d}{d\alpha}\left[e^{\alpha A^T} e^{\alpha A}\right]_{\alpha=0} = \frac{d}{d\alpha}[\mathbb{1}]_{\alpha=0} = 0

Applying the product rule on the left side:

\underbrace{\left[\frac{d}{d\alpha}e^{\alpha A^T}\right]_{\alpha=0}}_{A^T} \cdot \underbrace{e^{0 \cdot A}}_{\mathbb{1}} + \underbrace{e^{0 \cdot A^T}}_{\mathbb{1}} \cdot \underbrace{\left[\frac{d}{d\alpha}e^{\alpha A}\right]_{\alpha=0}}_{A} = A^T + A = 0

Therefore A^T = -A: the generators of \mathrm{SO}(n) are skew-symmetric (antisymmetric) matrices.

2.1) Counting Generators

An n \times n skew-symmetric matrix has \frac{n(n-1)}{2} independent entries (the upper triangle, with zeros on the diagonal):

Group	n	Number of generators \frac{n(n-1)}{2}
\mathrm{SO}(2)	2	1
\mathrm{SO}(3)	3	3
\mathrm{SO}(4)	4	6

The interactive graphic below lets you explore the skew-symmetric basis matrices for any \mathrm{SO}(n). Click on an entry or basis generator to see how it fits into the full matrix:

open graphic ↗

2.2) \mathrm{SO}(2): One Generator

\mathrm{SO}(2) has a single generator:

L = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

The Lie algebra is trivially [L, L] = 0 (a single generator always commutes with itself). The group elements are:

R(\alpha) = e^{\alpha L} = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix}

which are exactly the 2D rotation matrices.

2.3) \mathrm{SO}(3): Three Generators

The general 3 \times 3 skew-symmetric matrix decomposes as:

A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} = a \underbrace{\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}}_{A_1} + b \underbrace{\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}}_{A_2} + c \underbrace{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}}_{A_3}

Computing the commutators, we find [A_i, A_j] always yields \pm A_k, confirming closure.

To get cleaner commutation relations, we choose a different basis convention:

L_x = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \quad L_y = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \quad L_z = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

These satisfy the \mathrm{SO}(3) Lie algebra:

[L_i, L_j] = \epsilon_{ijk} \, L_k

where \epsilon_{ijk} is the Levi-Civita symbol (\epsilon_{xyz} = \epsilon_{yzx} = \epsilon_{zxy} = 1, \epsilon_{yxz} = \epsilon_{xzy} = \epsilon_{zyx} = -1, and = 0 if any indices repeat).

Axis-angle interpretation: The group element e^{\theta(\hat{v}_x L_x + \hat{v}_y L_y + \hat{v}_z L_z)} represents a rotation by angle \theta about the axis \hat{v} = (\hat{v}_x, \hat{v}_y, \hat{v}_z). The generators are intimately connected to the cross product: the matrix \hat{v}_x L_x + \hat{v}_y L_y + \hat{v}_z L_z acts on a vector \vec{w} by computing \hat{v} \times \vec{w}.

Rotations are in planes, not about axes. Looking at the basis generators, each one has nonzero entries in exactly one 2 \times 2 block — it generates rotation in a plane spanned by two coordinate axes (e.g. L_z rotates in the xy-plane). The "axis" picture is a special feature of 3D: there are \frac{3 \cdot 2}{2} = 3 planes and also 3 dimensions, so each plane has a unique perpendicular axis. In higher dimensions this coincidence breaks: \mathrm{SO}(4) has 6 generators (6 planes of rotation) in 4 dimensions — there is no unique "axis" for each rotation. The plane picture is the general one.

Note that the generators are not unique — similarity transforms can yield an equivalent basis. The Lie algebra (the commutation relations) is the invariant structure.

2.4) Group Multiplication and Closure Under Commutation

For a Lie group, group multiplication corresponds to combining exponentials. How does this work at the level of generators?

\mathrm{SO}(2): With a single generator, group multiplication is simple. Since [L, L] = 0, the BCH formula truncates immediately:

e^{\alpha L} e^{\alpha' L} = e^{(\alpha + \alpha') L}

Parameters simply add — this is why \mathrm{SO}(2) is abelian.

\mathrm{SO}(3): With three generators, the BCH formula involves commutators:

e^{\alpha_i A_i} e^{\alpha'_j A_j} = e^{\alpha_i A_i + \alpha'_j A_j + \frac{1}{2}[\alpha_i A_i, \alpha'_j A_j] + \ldots}

For this to be a valid group element, the exponent Z must be a linear combination of generators. The BCH series involves only commutators of the inputs — so this works if and only if the generators are closed under commutation:

[X_a, X_b] = f_{abc} \, X_c

Let's verify this explicitly for the A_i basis. Computing [A_1, A_2]:

[A_1, A_2] = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} = -A_3

Similarly, [A_2, A_3] = -A_1 and [A_3, A_1] = -A_2. The commutators close on the generators ✓

To remove the minus signs, we switch to the L_x, L_y, L_z basis (defined above), which gives the cleaner relation [L_i, L_j] = \epsilon_{ijk} L_k. The structure constants of \mathrm{SO}(3) in this basis are simply the Levi-Civita symbol \epsilon_{ijk}.

The key point: the commutation relations [X_a, X_b] = f_{abc} X_c are what define the Lie algebra. They encode the group's structure at the infinitesimal level. Two Lie groups with the same commutation relations (same structure constants) have the same local structure, even if they differ globally.

3) From Algorithms on Reps to Algorithms on Generators

We've seen how an infinite number of group elements can be generated from a finite set of generators via the exponential map. Can we use the generators directly — rather than the group elements — to do the things we've been doing for finite groups? Similarity transforms, direct sums, tensor products, decomposition into irreps?

The answer is yes. To see how, we need to know how common operations on representations translate to operations on generators. Here's the summary — derivations follow below:

Operation on reps	Operation on generators
Change of basis: Q D_1 = D_2 Q	Q X_a^{(1)} = X_a^{(2)} Q (same `infer_change_of_basis`!)
Similarity (special case): D' = U D U^{-1}	X'_a = U X_a U^{-1} (when Q is square & invertible)
Direct sum: D_1 \oplus D_2	X_a^{(1)} \oplus X_a^{(2)}
Tensor product: D_1 \otimes D_2	X_a \otimes \mathbb{1} + \mathbb{1} \otimes Y_a (Kronecker sum)

3.1) Change of Basis

For finite groups, we found matrices Q satisfying Q D_1(g) = D_2(g) Q using infer_change_of_basis. This works identically for Lie groups — just pass the generators instead of group elements:

Q X_a^{(1)} = X_a^{(2)} Q \qquad \text{for all generators } a

Why does this imply the same condition on all group elements? Starting from Q X_a = X'_a Q, we can show Q X_a^n = X_a'^n Q by induction. The key step:

Q X_a^2 = Q(X_a X_a) = (Q X_a) X_a = (X'_a Q) X_a = X'_a (Q X_a) = X'_a (X'_a Q) = X_a'^2 Q

Note that Q need not be invertible or even square — it can be rectangular (as when embedding a smaller irrep into a larger representation). The dimensions work out at each step: Q is d_1 \times d_2, X_a is d_2 \times d_2, and X'_a is d_1 \times d_1, so Q X_a^n is always d_1 \times d_2.

Applying this term by term to the exponential:

Q \, e^{\alpha X_a} = Q \sum_n \frac{\alpha^n}{n!} X_a^n = \sum_n \frac{\alpha^n}{n!} X_a'^n Q = e^{\alpha X'_a} \, Q

So the intertwining condition on generators implies the intertwining condition on all group elements. We only need to check finitely many generators instead of infinitely many group elements.

Change of basis on reps: Q D_1(g) = D_2(g) Q → On generators: Q X_a^{(1)} = X_a^{(2)} Q

When Q is square and invertible, this is a similarity transform: X'_a = Q X_a Q^{-1}.

3.2) Direct Sums

For a direct sum representation D(g) = D_1(g) \oplus D_2(g), the generator is the direct sum of generators:

D(g) = e^{\alpha_a X_a^{(1)}} \oplus e^{\alpha_a X_a^{(2)}} = e^{\alpha_a (X_a^{(1)} \oplus X_a^{(2)})}

This holds because the block-diagonal structure is preserved by powers: (A \oplus B)^n = A^n \oplus B^n (the blocks never mix). So:

X_a = X_a^{(1)} \oplus X_a^{(2)} = \begin{pmatrix} X_a^{(1)} & 0 \\ 0 & X_a^{(2)} \end{pmatrix}

The generators of a direct sum are the direct sum of the generators — exactly what you'd expect.

Direct sum on reps: D(g) = D_1(g) \oplus D_2(g) → On generators: X_a = X_a^{(1)} \oplus X_a^{(2)}

3.3) Tensor Products (Kronecker Sum)

For tensor products, something more interesting happens. The tensor product of representations is:

D(g) = D_1(g) \otimes D_2(g) = e^{\alpha_a X_a} \otimes e^{\alpha_a Y_a}

The generator of this tensor product representation is not the Kronecker product X_a \otimes Y_a — it's the Kronecker sum. To see why, expand both exponentials to first order in \epsilon:

(I_m + \epsilon X_a + \ldots) \otimes (I_n + \epsilon Y_a + \ldots) = I_m \otimes I_n + \epsilon(X_a \otimes I_n + I_m \otimes Y_a) + O(\epsilon^2)

The first-order term defines the generator of the tensor product:

Z_a = X_a \otimes \mathbb{1}_n + \mathbb{1}_m \otimes Y_a

This is called the Kronecker sum, written Z_a = X_a \overset{\text{kr}}{\oplus} Y_a. The full relationship is:

e^{\alpha_a X_a} \otimes e^{\alpha_a Y_a} = e^{\alpha_a(X_a \otimes \mathbb{1}_n + \mathbb{1}_m \otimes Y_a)}

Why the Kronecker sum? The Kronecker product of two exponentials e^A \otimes e^B equals e^{A \otimes I + I \otimes B} because A \otimes I and I \otimes B commute: [A \otimes I, I \otimes B] = (A \cdot I) \otimes (I \cdot B) - (I \cdot A) \otimes (B \cdot I) = A \otimes B - A \otimes B = 0. So the BCH formula gives simply e^{A \otimes I} e^{I \otimes B} = e^{A \otimes I + I \otimes B}.

Tensor product on reps: D(g) = D_1(g) \otimes D_2(g) → On generators: Z_a = X_a \otimes \mathbb{1} + \mathbb{1} \otimes Y_a (Kronecker sum)

All our decomposition machinery (infer_change_of_basis, irrep detection, CG coefficients) works for Lie groups: just stack constraints from the generators instead of from all group elements. Finitely many generators replaces infinitely many group elements.

4) Representations of \mathrm{SO}(2)

4.1) The Vector Representation is Reducible

The \mathrm{SO}(2) generator in the vector representation is:

L = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

This 2 \times 2 real matrix is reducible over the complex numbers. Finding its eigenvalues:

U L U^{-1} = i \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \qquad \text{with } U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}

So the complex-diagonal form is e^{\alpha \cdot i\sigma_z} = \begin{pmatrix} e^{i\alpha} & 0 \\ 0 & e^{-i\alpha} \end{pmatrix} = e^{i\alpha} \oplus e^{-i\alpha}.

The irreducible representations of \mathrm{SO}(2) are 1-dimensional and complex: e^{i\alpha} and e^{-i\alpha}.

4.2) All Irreps via Tensor Products

Since the irreps are 1D, the Kronecker sum for generators is just ordinary addition. Taking tensor products:

e^{i\alpha} \otimes e^{i\alpha} = e^{i2\alpha}, \qquad e^{-i\alpha} \otimes e^{-i\alpha} = e^{-i2\alpha}, \qquad e^{i\alpha} \otimes e^{-i\alpha} = e^{0 \cdot \alpha} = 1

All new irreps! The irreps of \mathrm{SO}(2) take the form e^{im\alpha} for integer m \in \mathbb{Z} (including m = 0, the trivial representation).

Selection rule: e^{im_1 \alpha} \otimes e^{im_2 \alpha} = e^{i(m_1 + m_2)\alpha}, so m_1 \otimes m_2 = m_1 + m_2. This is the simplest possible "CG decomposition" — tensor products of 1D irreps are always 1D.

Intuition: Objects that transform as the irrep e^{im\alpha} have m-fold rotational symmetry. In the real representation, we pair e^{im\alpha} and e^{-im\alpha} to form 2 \times 2 rotation matrices:

\begin{pmatrix} \cos(m\alpha) & \sin(m\alpha) \\ -\sin(m\alpha) & \cos(m\alpha) \end{pmatrix}

5) Representations of \mathrm{SO}(3)

5.1) The Vector Representation is Irreducible

Unlike \mathrm{SO}(2), the vector representation of \mathrm{SO}(3) (the three 3 \times 3 generators L_x, L_y, L_z) is irreducible. We can exponentiate each generator individually:

e^{\alpha L_x} = R_x(\alpha), \quad e^{\alpha L_y} = R_y(\alpha), \quad e^{\alpha L_z} = R_z(\alpha)

These are rotations in the yz-, xz-, and xy-planes respectively (or equivalently, in 3D only, rotations "about" the \hat{x}, \hat{y}, and \hat{z} axes). While each individual rotation matrix has a block structure (one coordinate is fixed, the other two rotate in a plane), the blocks do not align across generators — no single similarity transform can simultaneously block-diagonalize all three. This is the vector (or l = 1) irrep.

There is also the scalar representation (l = 0): L_x^{(0)} = L_y^{(0)} = L_z^{(0)} = 0 \Rightarrow e^0 = 1. This is unfaithful (all group elements map to the identity).

5.2) Building Irreps via Tensor Products

Starting from the l = 1 generators, we use tensor products and irrep decompositions to discover new irreps. This is the infer_irreps_from_tensor_products algorithm:

Keep two lists: p_todo (irreps to tensor product with the fundamental l = 1) and p_done (irreps already processed). Start with the scalar irrep l = 0 in p_todo.
Pop an irrep \rho from p_todo, tensor product it with l = 1 (using the Kronecker sum for generators), decompose using QS = RQ, check each resulting irrep for isomorphism with known irreps, and add any new ones to p_todo.
Repeat until we have as many irreps as desired.

Warning: The matrices get annoyingly large to do by hand very fast! Deriving irreps is best done computationally.

Why does this work? For compact Lie groups like \mathrm{SO}(n), the Peter–Weyl theorem guarantees that all irreps appear in the decomposition of sufficiently high tensor powers of the fundamental representation. This means repeated tensor products with the vector representation will eventually produce every irrep of the group. For further reading, see Georgi Ch. 2–3 and Hall, Lie Groups, Lie Algebras, and Representations, Ch. 12.

The first new irrep we discover is l = 2 (dimension 5). Its generators, obtained from the tensor product l{=}1 \otimes l{=}1 and decomposition, are:

L_x^{(2)} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 2 & 0 \end{pmatrix}

L_y^{(2)} = \begin{pmatrix} 0 & 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ -\sqrt{3} & 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \end{pmatrix}, \quad L_z^{(2)} = \begin{pmatrix} 0 & -\sqrt{3} & 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{pmatrix}

These are 5 \times 5 skew-symmetric matrices that satisfy the same Lie algebra [L_i^{(2)}, L_j^{(2)}] = \epsilon_{ijk} L_k^{(2)}. Note the \sqrt{3} entries — unlike l = 1 where all nonzero entries are \pm 1, higher irreps have entries that grow. The pattern of which entries are nonzero and their values are determined by the CG coefficients used in the tensor product decomposition.

The interactive graphic below lets you explore the generators, their squares, and the Casimir L^2 for any \mathrm{SO}(3) irrep up to l = 5:

open graphic ↗

5.3) The Casimir Operator L^2

Before computing irreps, we can see useful structure within any given irrep. Define the Casimir operator (or irrep invariant):

L^{(l)\,2} = L_x^{(l)\,2} + L_y^{(l)\,2} + L_z^{(l)\,2}

This operator commutes with all generators: [L^{(l)\,2}, L_i^{(l)}] = 0. By Schur's lemma, anything that commutes with all matrices of an irrep must be proportional to the identity:

L^{(l)\,2} = \lambda^{(l)} \, \mathbb{1}

l = 1 example: Computing each term explicitly:

L_x^2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad L_y^2 = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad L_z^2 = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}

Summing: L^{(1)\,2} = L_x^2 + L_y^2 + L_z^2 = -2 \cdot \mathbb{1}_3. Each individual L_i^2 is NOT proportional to the identity, but the sum is — as Schur's lemma requires.

l = 2 example: Using the 5 \times 5 generators from above:

L_x^{(2)\,2} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & 0 & -4 \end{pmatrix}, \quad L_y^{(2)\,2} = \begin{pmatrix} -3 & 0 & 0 & -\sqrt{3} & 0 \\ 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0 & 0 \\ -\sqrt{3} & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 \end{pmatrix}, \quad L_z^{(2)\,2} = \begin{pmatrix} -3 & 0 & 0 & \sqrt{3} & 0 \\ 0 & -4 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ \sqrt{3} & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 \end{pmatrix}

None of the individual L_i^{(2)\,2} are proportional to the identity — they have different diagonal entries and even off-diagonal elements. But their sum cancels perfectly:

L^{(2)\,2} = L_x^{(2)\,2} + L_y^{(2)\,2} + L_z^{(2)\,2} = -6 \cdot \mathbb{1}_5

More generally:

L^{(l)\,2} = -l(l+1) \, \mathbb{1}_{2l+1}

The eigenvalue -l(l+1) uniquely identifies the irrep. This is why the Casimir operator is useful: in a reducible representation (like a tensor product), computing L^2 reveals which irreps are present — the eigenvalues immediately tell you the l values.

l	\dim = 2l+1	L^2 = -l(l+1) \cdot \mathbb{1}
0	1	0
1	3	-2 \cdot \mathbb{1}_3
2	5	-6 \cdot \mathbb{1}_5
3	7	-12 \cdot \mathbb{1}_7

5.4) Diagonalizing by L_z: The \lvert l, m \rangle Basis

We cannot simultaneously block-diagonalize all three generators L_x, L_y, L_z (they don't commute). But we can choose to diagonalize one — conventionally L_z — and use its eigenvalues to label basis states within each irrep.

For l = 1, the unitary transform U that diagonalizes L_z is:

U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 1 \\ i & 0 & -i \\ 0 & \sqrt{2} & 0 \end{pmatrix}

Applying the similarity transform:

U^\dagger L_z^{(1)} U = \begin{pmatrix} -i & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & i \end{pmatrix} = L_z^{(1)\,\prime}

The eigenvalues are \{-i, 0, +i\}, corresponding to m = -1, 0, +1. We label the eigenstates as \lvert 1, m \rangle:

L_z^{(l)\,\prime} \lvert l, m \rangle = im \lvert l, m \rangle

Since L^{(l)\,2} commutes with L_z^{(l)}, we can simultaneously label states by eigenvalues of both L^2 (giving l) and L_z (giving m). This is the \lvert l, m \rangle basis favored by physicists.

The eigenvalues of L_z range from -l to l in integer steps, giving 2l + 1 states per irrep — consistent with the irreps being (2l+1)-dimensional.

Complex vs. real basis. Note that diagonalizing L_z produces complex generators and irrep matrices (the eigenvalues im are purely imaginary, and U contains i). This is the basis used in quantum mechanics and produces the complex spherical harmonics Y_l^m.

It is equally valid to work with the original real (skew-symmetric) generators without diagonalizing L_z. In this basis, the irrep matrices are real and the corresponding basis functions are the real spherical harmonics — the Cartesian monomials that appear in character tables (x, y, z, xy, x^2 - y^2, etc.). The real spherical harmonics do not have definite L_z eigenvalues; instead, each is a linear combination of the +m and -m complex harmonics.

In this course, we work in the real basis for computational simplicity. See the O(3) notes for more on conventions and the relationship between the two bases.

Ladder operators (for reference). In quantum mechanics, the range of m and the Casimir eigenvalue L^2 = -l(l+1) are traditionally derived using ladder operators L_\pm that shift m by \pm 1. The ladder operators annihilate the highest/lowest m states, which constrains m_\text{min} = -l and m_\text{max} = l. In this course we derive irreps computationally via tensor products and infer_change_of_basis, so ladder operators are not needed — but they provide useful intuition and are ubiquitous in the physics literature. See Griffiths QM 4.3 or Georgi Ch. 2 for the full derivation.

5.5) Tensor Product Decomposition and Selection Rules

Using the infer_irreps_from_tensor_products algorithm, we confirm computationally:

\underbrace{(l{=}1)}_{\text{3 dim}} \otimes \underbrace{(l{=}1)}_{\text{3 dim}} = \underbrace{(l{=}0)}_{\text{1 dim}} \oplus \underbrace{(l{=}1)}_{\text{3 dim}} \oplus \underbrace{(l{=}2)}_{\text{5 dim}}

Continuing: 1 \otimes 2 = 1 \oplus 2 \oplus 3, 1 \otimes 3 = 2 \oplus 3 \oplus 4, and so on.

The general selection rule (triangle inequality) for \mathrm{SO}(3) tensor products is:

l_1 \otimes l_2 = \lvert l_1 - l_2 \rvert \oplus (\lvert l_1 - l_2 \rvert + 1) \oplus \cdots \oplus (l_1 + l_2)

Equivalently, the irrep l_3 appears in l_1 \otimes l_2 if and only if:

\lvert l_1 - l_2 \rvert \leq l_3 \leq l_1 + l_2

This is the same triangle inequality that governs addition of angular momenta in quantum mechanics.

5.6) Clebsch-Gordan Coefficients

The Clebsch-Gordan (CG) coefficients are the change-of-basis matrices that decompose a tensor product representation into irreps — i.e., they are solutions to QS = RQ where S is the tensor product representation and R is an irrep.

The procedure:

Tensor product the generators (Kronecker sum)
Decompose into irreps using QS = RQ
The resulting change-of-basis matrix Q contains the CG coefficients
Reshape and transpose into the standard form C_{l_1 m_1, l_2 m_2}^{l_3 m_3}

By Schur's lemma, we can always multiply an irrep by a constant matrix and get the same irrep. This means we can choose to normalize the CG coefficients however we want — there are many conventions. In the symm4ml library and in e3nn, we choose normalizations that favor maintaining standard deviations of 1 when building neural networks from these operations.

Can also use CG coefficients to derive basis functions that transform as irreps of \mathrm{SO}(3) \rightarrow spherical harmonics.

6) Resources

Georgi, Lie Algebras in Particle Physics, Chapter 2 — introduction to Lie groups
Georgi, Chapter 6 — \mathrm{SO}(3) irreps and tensor methods
Griffiths, Introduction to Quantum Mechanics, Section 4.3 — angular momentum and ladder operators
Course notes on representations — Schur's lemma and the intertwining equation for finite groups
Course notes on decomposition — tensor product decomposition algorithm